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A particle is projected from ground at a...

A particle is projected from ground at angle `45^@` with initial velocity `20(sqrt2) m//s`. Find
(a)change in velocity,
(b)magnitude of average velocity in a time interval from `t=0` to `t=3s`.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
(a) Since, acceleration is constant. Therefore,
`a_(av)= a=g=(-10hatj) = (Deltav)/(Deltat)`
`:. Delta v = (-10hatj)(Deltat) = (-10hatj)(3)`
`= (-10 hatj)(3) = (-30hatj)m//s`
`:.` Change in velocity is `30 m//s`, vertically downwards.
(b)`v_(av) = s/t = (ut + 1/2at^2)/t = u+1/2at.`
`=(20hati + 20hatj) + 1/2 (-10hatj)(3)`
`=(20hati + 5hatj) m//s`
`:. |v_(av)| = (sqrt (20)^2+(5)^2)`
`=20.62 m//s`
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